Summations Question I'm not used to dealing with summations and there's a lot of summation in probability apparently. There's all these tips and technique that the books assume you to know. I'm used to working with integrals. So are there summations/summation techniques I should familiarize myself with.
So far 
I have encountered binomial, geometric and  telescoping.
 A: This isn't an exact method but, to approximate sums of a given monotonic function $f$ over the interval $(a,b)$, you can say $$\int_{a-1}^{b}f(x)\text dx\le \sum_{k=a}^bf(k)\le \int_a^{b+1} f(x)\text dx$$
if $f$ is increasing and 
$$\int_{a-1}^{b}f(x)\text dx\ge \sum_{k=a}^bf(k)\ge \int_a^{b+1} f(x)\text dx$$
if $f$ is decreasing.
Sometimes you get lucky enough to use Squeeze Theorem to evaluate the sum, like here:
$$\begin{align}\lim_{n\to\infty}\int_1^{n+1}\frac{1}{x+n}\text dx\le\lim_{n\to\infty}\sum_{k=1}^n \frac{1}{k+n}&\le \lim_{n\to\infty}\int_0^{n}\frac{1}{x+n}\text dx\\
\lim_{n\to\infty}\log(2n+1)-\log(n+1)\le\lim_{n\to\infty}\sum_{k=1}^n \frac{1}{k+n}&\le\lim_{n\to\infty}\log2\\
\log 2\le\lim_{n\to\infty}\sum_{k=1}^n\frac1{k+n} &\le \log 2\\
\lim_{n\to\infty}\sum_{k=1}^n\frac1{k+n}&=\log 2\end{align}$$

Every once in a while, you can also use $$\lim_{n\to\infty}\frac1n\sum_{k=an}^{bn}f\left(\frac kn\right)=\int_a^bf(x)\text dx$$
The adept eye would notice that you can use this in the example I gave above.

Also Maclaurin series can be very useful for computing infinite sums, i.e.
$$ f(x-n)=\sum_{k=0}^\infty f^{(k)}(n)\frac{(x-n)^k}{k!} $$
Where $f^{(k)}(n)$ is the $k^{th}$ derivative of $f$ at $n$.
For example, $$ 1+\sum_{k=1}^\infty \frac{\prod_{i=1}^k (3i-1)}{3^k k!}2^{-k} $$ can be found to evaluate to $\sqrt[3]4$ with the Maclaurin series of $x^{2/3}$ about $x=1$.

One other useful trick can be to take the derivative or integral of the sum to get it into a nicer form.  For example, $$\sum_{k=0}^{n-1}x^k(k+1)=\frac{\text d}{\text dx}\sum_{k=0}^nx^k=\frac{\text d}{\text dx}\frac{x^{n+1}-1}{x-1}=\frac{nx^{n+1}-(n+1)x^n+1}{(x-1)^2}$$
This last one can also get useful identities like, for $|x|<1$, $$-\log(1-x)=\int\frac{1}{1-x}\text dx=\int\sum_{k=0}^{\infty}x^k\text dx=\sum_{k=1}^\infty \frac{x^k}{k}$$
A: If you really want to know, there is the book by D Knuth (and coauthors)
concrete mathematics:
http://www.amazon.com/Concrete-Mathematics-Foundation-Computer-Science/dp/0201558025
